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I am new to proof writing. If trying to prove for example: $$(ab)^{-1} = a^{-1} b^{-1}$$

What am I trying to do - I either pick the left or right side as my starting point and manipulate it algebraically in isolation of the opposing side until I derive the opposing side? It makes no difference if I choose the left or right side of the equation as my starting point?

lopan
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    That'll be hard to prove, since it's false for nonabelian groups – Alan Jul 12 '21 at 00:13
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    Thanks for the response, but I think my question was so basic that you may have missed it. – lopan Jul 12 '21 at 00:15
  • What are $a$ and $b$ elements of? – Shaun Jul 12 '21 at 00:15
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    Yes, it would be valid to start trying to algebraically manipulate the left hand side into the right hand side or visa versa the right to the left. To piggy back on @Alan 's comment, we presume you are trying to prove the basic group theoretic result $\forall a,b \in G: (ab)^{-1} = b^{-1}a^{-1} $. The result you are trying to prove is false in the general case if you are indded working with groups! – user2628206 Jul 12 '21 at 00:15
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    There is no difference as to which side you begin with. A supporting argument for this is that equality is a symmetric relation, i.e. $a=b$ is the same as $b=a$. Your algebraic manipulations are an application of the transitivity of equality, that is, it follows from $a=b$ and $b=c$ that $a=c$. – Denis Gorodkov Jul 12 '21 at 00:16
  • Once I begin the proof, I need to have discarded the side of the equation that I don't choose to begin with right? In other words I can't drag over terms from the other side of the = during the proof right? – lopan Jul 12 '21 at 00:19
  • The left hand side is, by definition, the multiplicative inverse of $ab$. Show that the right-hand side satisfies that definition as well, and it follows that the two must be equal. – Matthew Leingang Jul 12 '21 at 00:20
  • I'm not quite sure what you are meaning, but a heuristic peace of advice would be to once you have picked a side to manipulate, don't appeal to the other side until you have proved your result. (so no dragging terms). @MatthewLeingang has given the winning strategy for solving this problem of groups. – user2628206 Jul 12 '21 at 00:21
  • @lopan Suppose you want to prove $a=b$ (that this statement is true). If you know that $a=b$, $b=c$ and $c=d$ are all true, then transitivity of equality gives you the answer that $a=d$ is indeed true. – Denis Gorodkov Jul 12 '21 at 00:35
  • Thanks @user2628206, that makes sense. – lopan Jul 12 '21 at 00:38

1 Answers1

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I'm going to assume that $a$ and $b$ are non-zero real numbers (for simplicity).

In any proof, we always begin by checking definitions. In this case, the definition of $x^{-1}$ is: the unique value $y$ such that $x \cdot y = 1$.

So we know that $(ab) \cdot (ab)^{-1} = 1$. We also know that

$\begin{equation} \begin{split} (ab) \cdot (a^{-1} \cdot b^{-1}) &= ((a \cdot b) \cdot a^{-1}) \cdot b^{-1} \\ &= ((b \cdot a) \cdot a^{-1}) \cdot b^{-1} \\ &= (b \cdot (a \cdot a^{-1})) \cdot b^{-1} \\ &= (b \cdot 1) \cdot b^{-1} \\ &= b \cdot b^{-1} \\ &= 1 \end{split} \end{equation} $

So since $(ab)^{-1}$ is the only $y$ such that $(ab) \cdot y = 1$, and it is also true that $(ab) \cdot (a^{-1} b^{-1}) = 1$, it must be the case that $(ab)^{-1} = a^{-1} b^{-1}$.

Notice that we always start with the definition. Once you know the definition of a thing, you can begin to derive its properties.

Mark Saving
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  • Great thanks. When you say in general to "always begin by 'checking' definitions", what precisely do you mean by 'checking'? – lopan Jul 12 '21 at 00:47
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    @lopan "checking" here just means read and make sure you know the definitions (ideally you should understand the definition on a basic level... though this is easier said than done the more advanced things get), because without knowing definitions, you cannot even begin to prove anything. Afterall, if one does not know what they are talking about, how can they ever say anything useful/interesting about it :) Here, we're checking to see that the quantity $a^{-1}b^{-1}$ satisfies the definition of $(ab)^{-1}$. – peek-a-boo Jul 12 '21 at 00:55
  • I mistakenly thought my goal in writing a proof was simply to show how to derive one side of the equation from the other. Is the correct idea that instead, I am trying to demonstrate that both sides of the equation are an expression of the definition, where in this case the definition is as @Mark Saving lays it out above? If so, in this particular proof, is the idea that the left side is obviously an expression of the definition, but the right side isn't, so here I would only have to show how the right side is an expression of the definition? – lopan Jul 12 '21 at 01:42
  • @Iopan Yes, that is the idea. It is often the case that you can prove an equality by applying a series of "rewrite steps", as you suggest, rather than working from the definitions directly. In this case, we can't do that directly - we're trying to prove that an elementary rewriting step is valid, so we don't have any tools to rely on except the definition. – Mark Saving Jul 12 '21 at 01:58
  • Awesome, thanks for clarifying that distinction. – lopan Jul 12 '21 at 02:35
  • @MarkSaving While I agree with your proof approach, how exactly you prove this depends on what else you know. In your case you take it as written you have commutativity and associativity as axioms, or theorems. But even so if this showed up in my class as a (basic) proof it would likely get very few points. If you are at a point where you are proving something like this for the reals I would absolutely require a jusitification for every equals line. – DRF Jul 12 '21 at 08:35
  • @DRF I agree. I must confess that the only reason I did not explicitly justify each line is because I forgot how to format it nicely in LaTeX, but hopefully it's clear that each line cites exactly one "basic rule" of associativity, commutativity, right identity, and the definition of inverses. – Mark Saving Jul 12 '21 at 15:40